
\begin{section}{The Craighero-Gattazzo Surface}

\begin{subsection}{Introduction}

The Craighero-Gattazzo (CG) surface is an algebraic surface in $\mathbb{P}^{3}$ described by a quintic polynomial $\mathcal{F}_{5}$ in four indeterminants $x,y,z,t$ and six coefficients $a,b,c,e,f,m$.

By construction, the CG surface has four $\tilde{E}_{8}$ singularities invariant under the $\mathbb{Z}_{4}$ action on $\mathbb{P}^{3}$. This requirement forces the following restrictions on coefficients:

\begin{equation} \label{eq:cg1}
g_{1} = -a^5 + a^2e - ab + 1 = 0.
\end{equation}

\begin{equation} \label{eq:cg2}
g_{2} = -3a^4m + a^2b + 2aem - af - bm + c = 0.
\end{equation}

\begin{equation} \label{eq:cg3}
g_{3} = -3a^3m^2 + 2abm + cm^2 - ac - fm + c = 0.
\end{equation}

\begin{equation} \label{eq:cg4}
g_{4} = -9a^8 + 12a^5e - 6a^4b - 8a^4m - 4a^2e^2 + 4a^2c + 4abe - b^2 - 8am = 0.
\end{equation}

\begin{equation} \label{eq:cg5}
g_{5} = mb + 3a^3mb - abc + a^2ce - 4a^4me - a^2m^2 + a^3m^2 - 3a^4m + 3a^7m = 0.
\end{equation}

\begin{equation} \label{eq:cg6}
\begin{split}
g_{6} & = 18a^7m - 6a^5b - 18a^4em + 3a^4f + 6a^3bm - 6a^3m^2 - 2a^4 + 4a^2be + 4ae^2m \\ 
& - 2ab^2 + a^2f - 2aef + 2aem - 2bem - af + bf -2m^2 + 2a = 0.
\end{split}
\end{equation}

The coefficients are defined as follows:

\begin{equation} \label{eq:a1}
a = r^2.
\end{equation}

\begin{equation} \label{eq:b1}
b = -\frac{1}{7}(2r^2 - 13r - 18).
\end{equation}

\begin{equation} \label{eq:c1}
c = \frac{1}{49}(73r^2 + 75r + 92).
\end{equation}

\begin{equation} \label{eq:e1}
e = -\frac{1}{7}(r^2 - 24r - 9).
\end{equation}

\begin{equation} \label{eq:f1}
f = \frac{1}{49}(181r^2 + 241r + 163).
\end{equation}

\begin{equation} \label{eq:m1}
m = \frac{1}{7}(3r^2 + 5r + 1).
\end{equation}

Where $r$ is a root of the cubic polynomial $t^3 + t^2 - 1$.

\end{subsection}

\begin{subsection}{Deriving the Restrictions on Coefficients}

\begin{subsubsection}{The First Three Restrictions}

Let $\mathcal{F}$ be the equation for the quintic. After dehomogenizing with $t = 1$ and making the quadratic part into the square of a linear term, we get

\begin{equation}
\mathcal{F}_{3} = \alpha y^3 + \beta y^{2}z + \delta yz^{2} + \gamma z^3.
\end{equation}

We want to have 
$$\mathcal{F}_{3} = \alpha y^3  = (-a^2m^3 + bm^2 - cm + m^2)y^3,$$ 
which requires $\beta = \delta = \gamma = 0$ and places the following three restrictions on the coefficients of $\mathcal{F}$.

\begin{equation}
\boxed{g_{1} = -a^5 + a^2e - ab + 1 = 0.}
\end{equation}

\begin{equation}
\boxed{g_{2} = -3a^4m + a^2b + 2aem - af - bm + c = 0.}
\end{equation}

\begin{equation}
\boxed{g_{3} = -3a^3m^2 + 2abm + cm^2 - ac - fm + c = 0.}
\end{equation}

\begin{rmk}
Note that
\begin{equation}
\begin{split}
g_{1} & = 0 \iff \gamma = 0,\\
g_{2} & = 0 \iff \delta = 0,\\
g_{3} & = 0 \iff \beta = 0.
\end{split}
\end{equation}
\end{rmk}
\end{subsubsection}

\begin{subsubsection}{The Fourth Restriction}
Then we do a blowup of the surface in the direction of $z$, using the quadratic change of coordinates $$(x,y,z) \to (xz, yz, z)/z^2.$$

After the blowup, we examine $\mathcal{F}_{2}$ and notice that the coefficient of the $yz$ term is exactly $\delta = 0$, which vanishes and allows us to write

\begin{equation}
\mathcal{F}_{2} = (x + \mu z)^2 = x^2 + 2\mu xz + \mu^2 z^2.
\end{equation}

A quick calculation shows that $$\mu = \frac{3a^{4} - 2ae + b}{2}$$.

In order to make $\mathcal{F}_{2} = 0$, we need to change coordinates by $(x,z) \to (-z\mu, z)$. This imposes a fourth restriction on the coefficients of $\mathcal{F}$. That is,

\begin{equation}
\boxed{g_{4} = -9a^8 + 12a^5e - 6a^4b - 8a^4m - 4a^2e^2 + 4a^2c + 4abe - b^2 - 8am = 0.}
\end{equation}

\begin{rmk}
Note that $g_{4} = 0 \iff \nabla = 0$, where $\nabla$ is the ``discriminant'' of $\mathcal{F}_{2}$.
\end{rmk}
\end{subsubsection}

\begin{subsubsection}{The Fifth and Sixth Restrictions}
Now after changing coordinates $(x,z) \to (-z\mu, z)$ from the last step, we have $$\mathcal{F}_{3} = \tilde{\alpha}z^3 + \tilde{\beta}yz^2 + \tilde{\delta}y^2z.$$

The coefficients of $\mathcal{F}_{3}$ are:

\begin{equation}
\begin{split}
\tilde{\alpha} & = -27a^{11} + 45a^7e - 18a^7m - 24a^5e^2 - 6a^5c + 18a^4be - 12a^4em - 3a^3b^2 + 4a^2e^3 + 6a^4m \\
& + 6a^3bm - a^3m^2 + 4a^2ce - 4abe^2 + a^2m^2 - 2abc + b^2e - 4aem + 2bm.
\end{split}
\end{equation}

\begin{equation}
\begin{split}
\tilde{\beta} & = 18a^7m - 6a^5b - 18a^4em + 3a^4f + 6a^3bm - 6a^3m^2 - 2a^4 + 4a^2be + 4ae^2m - 2ab^2 + a^2f \\
& - 2aef + 2aem - 2bem - af + bf -2m^2 + 2a.
\end{split}
\end{equation}

\begin{equation}
\tilde{\delta} = -3a^3m^2 + 2abm + em^2 - ac - fm + c.
\end{equation}

Note that $\tilde{\delta} = \delta = 0$ causes the $y^2z$ term to vanish, leaving us with $\mathcal{F}_{3} = \tilde{\alpha}z^3 + \tilde{\beta}yz^2$. Now we want to have $\mathcal{F}_{3} = 0$, which requires two further restrictions on the coefficients of $\mathcal{F}$.

\begin{prop}
Let $\lambda = -\frac{1}{6}\tilde{\alpha}$. Then
\begin{equation}
\frac{g_{4}(3a^3 - e) + \lambda}{4} = 0 \iff \tilde{\alpha} = 0.
\end{equation}
Hence, $$g_{5} = \frac{g_{4}(3a^3 - e) + \lambda}{4}.$$
\end{prop} 

That is, the requirement for $\tilde{\alpha} = 0$ can be expressed more simply (involving a polynomial of smaller degree with fewer terms) by setting $g_{5} = 0$. Explicitly, we have

\begin{equation}
\boxed{g_{5} = mb + 3a^3mb - abc + a^2ce - 4a^4me - a^2m^2 + a^3m^2 - 3a^4m + 3a^7m = 0.}
\end{equation}

The sixth and final restriction comes directly from setting $\tilde{\beta} = 0$.

\begin{equation}
\boxed{\begin{split}
g_{6} & = 18a^7m - 6a^5b - 18a^4em + 3a^4f + 6a^3bm - 6a^3m^2 - 2a^4 + 4a^2be + 4ae^2m - 2ab^2 + a^2f \\
 & - 2aef + 2aem - 2bem - af + bf -2m^2 + 2a = 0.
\end{split}}
\end{equation}

\end{subsubsection}

\end{subsection}

\end{section}
